# Solve the following inequality $x^2+x+1\gt 0$

Solve the following inequality $x^2+x+1\gt 0$

I understand how to solve inequalities and what the graphs look like. Usually the first step is to set this as in equation and then find the zeros. But for this one when I used the quadratic formula my two answers were: $$\dfrac{-1+i\sqrt{3}}{2}$$ $$\dfrac{-1-i\sqrt{3}}{2}$$

I do not know what to do after this, or if I messed up in any way. Or, if there is another way to solve this problem. Any hints help. Please do not solve this problem in any way for me. Thanks!

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$\sqrt{-3}\neq\sqrt{3i}$. –  Cameron Buie Jul 20 '12 at 6:19
I had it right on my paper! Thanks for the edit. I didn't catch it. –  Austin Broussard Jul 20 '12 at 6:20
Can you simplify $x^2+x+1$ by completing the square? –  Martin Sleziak Jul 20 '12 at 6:20
I really dislike completing the square because nobody has yet to explain it to me in a way that I understand –  Austin Broussard Jul 20 '12 at 6:20

Complete the square: $x^2+x+1=\left(x+\frac12\right)^2+\frac34$. For what values of $x$ is this positive?

Added: Here’s an explanation of completing the square. Suppose that you have a quadratic $x^2+ax+b$. You want to write this in the form $(x+c)^2+d$ for some constants $c$ and $d$. We know that $(x+c)^2+d=x^2+2cx+(c^2+d)$, and we want this to be identically equal to $x^2+ax+b$. That is, we want $x^2+2cx+(c^2+d)$ and $x^2+ax+b$ to be the same polynomial. Clearly this means that we must have $2c=a$ and $c^2+d=b$. In particular, we must have $c=\frac{a}2$. I could also solve for $d$, but in practice it’s easier to work it out each time than it is to use a formula.

Knowing now that $c=\frac{a}2$, I write $\left(x+\frac{a}2\right)^2$ as a first approximation to $x^2+ax+b$, and then I multiply it out to get $x^2+ax+\frac{a^2}4$. This approximation gives me the right $x^2$ and $x$ terms, but in general it gives me the wrong constant term, because $\frac{a^2}4$ is rarely equal to $b$. Therefore I have to adjust my approximation $\left(x+\frac{a}2\right)^2$. I do so by subtracting $\frac{a^2}4$ and adding $b$:

$$\left(x+\frac{a}2\right)^2-\frac{a^2}4+b=x^2+ax+b\;,$$

as desired. In the case of the quadratic $x^2+x+1$, $a=1$, so $c=\frac{a}2=\frac12$, and my first approximation was $\left(x+\frac12\right)^2$. This has a constant term of $\frac14$ instead of the desired $1$, so I knew that I had to add another $\frac34$.

The only remaining issue is what to do when the coefficient of $x^2$ isn’t $1$, i.e., when we’re dealing with $ax^2+bx+c$ with $a\ne 1$. The easiest approach is to factor out the $a$ to get

$$a\left(x^2+\frac{b}ax+\frac{c}a\right)\;.$$

Then complete the square on $x^2+\frac{b}ax+\frac{c}a$ to get

$$a\left(\left(x+\frac{b}{2a}\right)^2+\text{some constant}\right)\;.$$

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Considering all values because when $(0+\frac{1}{2})^2+\frac{3}{4}=1$ and every number, negative and positive, are all positive and they increase. –  Austin Broussard Jul 20 '12 at 6:25
@Austin: Not quite: the minimum value is $\frac34$, when $x=-\frac12$. But what you can say is that $\left(x+\frac12\right)^2\ge 0$, so $x^2+x+1\ge\frac34>0$ for all $x$. –  Brian M. Scott Jul 20 '12 at 6:27
Okay, so $\frac{3}{4}$ is the minimum value and does satisfy the OP, right? –  Austin Broussard Jul 20 '12 at 6:31
@Austin: Yes, it does. And having now seen your comment about completing the square, I’ve added an explanation of the process in hope of making this particular tool more readily available to you in the future. –  Brian M. Scott Jul 20 '12 at 6:42
@Austin: You could write the solution set in many ways; some of them are $\Bbb R$, $\{x:x\in\Bbb R\}$, $(-\infty,\infty)$, $\text{all real numbers}$, $\text{all }x\in\Bbb R$. –  Brian M. Scott Jul 20 '12 at 7:08

It has no real zeroes and is continuous. What does this allow you to conclude about the sign of the function?

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Why don't you try plotting the graph of the function and guess?

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Because I wanted a better understanding. Not just graph it and pull the answer from there. I need a definite answer, not a guess. –  Austin Broussard Jul 20 '12 at 7:13
Well, from the graph, it is clear there are no real zeros. This means it doesn't change sign. Setting $x=0$ gives $1$ which means the inequality is satisfied everywhere. You could also differentiate and find the minimizing $x=\frac{1}{2}$, and value at this point $\frac{7}{4}$. I'm not suggesting that you only graph it and guess, I'm suggesting that you graph it to understand why solving the quadratic produces 'unreal' values. –  copper.hat Jul 20 '12 at 7:16